hdu.1097.A hard puzzle

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41771    Accepted Submission(s): 15012

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

Output

For each test case, you should output the a^b's last digit number.

 

Sample Input

7 668 800

 

Sample Output

96
这道题第一眼看觉得比较简单，但按照常规的求，肯定会超时。存在以下规律：
尾数
1 :     1 1 1 1 1 1 1 1
2 :     2 4 8 6 2 4 8 6
3 :     3 9 7 1 3 9 7 1
4 :     4 6 4 6 4 6 4 6
7 :     7 9 3 1 7 9 3 1
可以看出来四个一个循环

#include <stdio.h>#include <stdlib.h>int main(){    int a,b,i,s;    while(scanf("%d%d",&a,&b)==2)    {        s=1;        a=a%10;        b=b%4;        if(b==0)        {            b=4;        }        for(i=0;i<b;i++)        {            s=s*a;            s=s%10;        }        printf("%d\n",s);    }    return 0;}